Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
#include<unordered_map>
#pragma warning(disable:4996)
using namespace std;

const int My_Inf=2147483647;
const long long int My_LInf=9223372036854775807;

struct aa {
	int to;
	int time;
};
class Compare {
public:
	//keiroが昇順に並ぶ
	bool operator()(const aa&l, const aa&r) {
		return l.time> r.time;
	}
};

int main() {
	int N, M, S, T; cin >> N >> M >> S >> T;
	S--; T--;
	vector<vector<int>>edges(N);
	for (int i = 0; i < M; ++i) {
		int f, t; cin >> f >> t;
		f--; t--;
		edges[f].push_back(t);
		edges[t].push_back(f);
	}

	vector<int>starttimes(N, 999499);
	vector<long long int >starttimenum(1100000);
	{
		priority_queue<aa, vector<aa>, Compare>que;
		que.push(aa{ S,0 });
		while (!que.empty()) {
			aa atop(que.top());
			que.pop();
			starttimes[atop.to] = atop.time;
			for (auto i : edges[atop.to]) {
				if (starttimes[i] > atop.time + 1) {
					starttimes[i] = atop.time + 1;
					que.push(aa{ i,atop.time + 1 });
				}
			}
		}
		for (int i = 0; i < starttimes.size(); ++i) {
			starttimenum[starttimes[i]]++;
		}
	}
	vector<int>goaltimes(N, 999499);
	vector<long long int >goaltimenum(1100000);
	{
		priority_queue<aa, vector<aa>, Compare>que;
		que.push(aa{ T,0 });
		while (!que.empty()) {
			aa atop(que.top());
			que.pop();
			goaltimes[atop.to] = atop.time;
			for (auto i : edges[atop.to]) {
				if (goaltimes[i] > atop.time + 1) {
					goaltimes[i] = atop.time + 1;
					que.push(aa{ i,atop.time + 1 });
				}
			}
		}
		for (int i = 0; i < goaltimes.size(); ++i) {
			goaltimenum[goaltimes[i]]++;
		}
	}

	int fintime = starttimes[T];

	long long int ans = 0;
	for (int i = 0; i < fintime-1; ++i) {
		ans += starttimenum[i] * goaltimenum[fintime-i-2];
	}
	cout << ans << endl;
	return 0;
}